Question

A metal forms two oxides. The higher oxide contains `80 %` metal. `0.72 g` of the lower oxide have `0.8 g` of higher oxide when oxidised. Show that the data illustrates the law of Multiple Proportions.

Answer 1

In the higher oxide :
Weight of metal `= ((0.8g)xx80)/(100)=0.64g`
Weight of oxygen `=0.80-0.64=0.16g`
In the lower oxide :
Weight of metal will remain the same as in the higher oxide and is `0.64 g`
`:.` Weight of oxygen `= 0.72 - 0.64 = 0.08g`
The ratio of weights of oxygen which combine with a fixed weight of metal `(0.64 g)` in the two oxides is :
`0.16 : 0.08 or 2:1`
Since the ratio is simple whole number in nature, the law Multiple proportions is illustrated.